What are the divisors of 8295?

1, 3, 5, 7, 15, 21, 35, 79, 105, 237, 395, 553, 1185, 1659, 2765, 8295

There is a total of 16 positive divisors.
The sum of these divisors is 15360.
The arithmetic mean is 960.

16 odd divisors

1, 3, 5, 7, 15, 21, 35, 79, 105, 237, 395, 553, 1185, 1659, 2765, 8295

How to compute the divisors of 8295?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 8295 by each of the numbers from 1 to 8295 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

8295 / 1 = 8295 (the remainder is 0, so 1 is a divisor of 8295)
8295 / 2 = 4147.5 (the remainder is 1, so 2 is not a divisor of 8295)
8295 / 3 = 2765 (the remainder is 0, so 3 is a divisor of 8295)
...
8295 / 8294 = 1.0001205690861 (the remainder is 1, so 8294 is not a divisor of 8295)
8295 / 8295 = 1 (the remainder is 0, so 8295 is a divisor of 8295)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 8295 (i.e. 91.07689059251). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

8295 / 1 = 8295 (the remainder is 0, so 1 and 8295 are divisors of 8295)
8295 / 2 = 4147.5 (the remainder is 1, so 2 is not a divisor of 8295)
8295 / 3 = 2765 (the remainder is 0, so 3 and 2765 are divisors of 8295)
...
8295 / 90 = 92.166666666667 (the remainder is 15, so 90 is not a divisor of 8295)
8295 / 91 = 91.153846153846 (the remainder is 14, so 91 is not a divisor of 8295)